# Exercise with Square Spiral

### What Is This About?

### Source

Hubert Shutrick has kindly communicated to me his investigation of a hand-drawn spiral, as bellow:

This is a simple construction that leads to a self-similar shape, concerning which several questions may be meaningfully asked as to the lengths and areas involved. Answers to these questions reveal that, the self-similarity notwithstanding, neither the boundary curves, nor the enclosed form, are fractals - just plain self-similar figures. The curves have finite lengths, whereas the enclosed shape has finite area. There are certainly weirder cases.

### Questions

There are several questions that may be asked concerning the resulting fractal. The shape consists of quarter circles stuck together. It is bounded by two spiral curves: the smooth outer spiral and the rectilinear inner spiral.

- What is the length of the smooth outer spiral?
- What is the length of the rectilinear inner spiral?
- What is the area enclosed between the two curves?
- What is the accumulation point of the two spirals?

### Answers

For all questions the answer comes from a formula for a sum of the geometric series:

$\displaystyle a+aq+aq^2+aq^3+\ldots =a\frac{1}{1-q}=\frac{a}{1-q}.$

**What is the length of the smooth outer spiral?**$\displaystyle a=\frac{2\pi}{4}=\frac{\pi}{2},\,$ $\displaystyle q=\frac{1}{2},\,$ $\displaystyle\frac{a}{1-q}=\pi.$

**What is the length of the rectilinear inner spiral?**$a=1,\,$ $\displaystyle q=\frac{1}{2},\,$ $\displaystyle\frac{a}{1-q}=2.$

**What is the area enclosed between the two curves?**$\displaystyle a=\frac{\pi}{4},\,$ $\displaystyle q=\frac{1}{4},\,$ $\displaystyle\frac{a}{1-q}=\frac{\pi}{3}.$

**What is the accumulation point of the two spirals?**If the accumulation point - the tip of the spiral shape - is described as $(x_0,y_0),\,$ we'll compute the two separately:

$x_0=\displaystyle\frac{1}{4}-\frac{1}{16}+\frac{1}{64}-\ldots,\;$ so that $a=\displaystyle\frac{1}{4},\,$ $q=\displaystyle -\frac{1}{4}.\;$ Thus,

$\displaystyle\frac{a}{1-q}=\frac{1}{4}\cdot\frac{4}{5}=\frac{1}{5}.$

For $y_0,\,$ the calculations are the same, except $\displaystyle a=\frac{1}{2},\,$ so that $\displaystyle\frac{a}{1-q}=\frac{1}{2}\cdot\frac{4}{5}=\frac{2}{5}.$

We come up with the accumulation point $\displaystyle (x_0,y_0)=\left(\frac{1}{5},\frac{2}{5}\right).$

### Answers using self-similarity

Hubert Shutrick's original solutions dependent on the shape at hend being self-similar. The first piece aside, the rest of the shape is constructed exactly like the whole but strating with half as large quarter circle (and this rotated $90^{\circ}\,$ degrees clockwise). This remark leads to a few simple equations:

**What is the length of the smooth outer spiral?**If the sought length is $L\,$ then $\displaystyle \frac{\pi}{2}+\frac{L}{2}=L,\,$ implying $L=\pi.$

**What is the length of the rectilinear inner spiral?**If the sought length is $M\,$ then $\displaystyle 1+\frac{M}{2}=M,\,$ implying $M=2.$

**What is the area enclosed between the two curves?**If $S\,$ is the said area, $\displaystyle S=\frac{\pi}{4}+\frac{1}{4}S,\,$ implying $S=\displaystyle\frac{\pi}{3}.$

**What is the accumulation point of the two spirals?**If $z\,$ is the accumulation we have $\displaystyle z=\frac{i}{2}-\frac{i}{2}z,\,$ implying

$\displaystyle\begin{align} 2z&=i-iz,\\ (2+i)z&=i,\\ z&=\frac{i}{2+i}=\frac{i(2-i)}{5}=\frac{1}{5}+i\frac{2}{5}. \end{align}$

It's possible to avoid complex variables by observing that $\displaystyle\left(\frac{3}{16},\frac{3}{8}\right),$ is the new origin of a copy size $\displaystyle\frac{1}{16}-\text{th}\,$ so the vector $v\,$ from the origin to the accumulation point satisfies

$\displaystyle v = \left(\frac{3}{16},\frac{3}{8}\right) + \frac{v}{16}.$

Finally, Hubert also observed that the homothety from the whole elephant to the one $\displaystyle\frac{1}{16}-\text{th}\,$ its size is a projection from the accumulation point which can therefore be also found as the intersection of the line of centres $y=2x\,$ and the line of starting points of the spirals $3y-x=1.$

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